Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m, q;
int in[100000 + 5], out[100000 + 5], t = 0;
bool vis[100000 + 5];
vector<int> eg[100000 + 5];
bool uisa(int u, int v) { return in[u] <= in[v] && out[u] >= out[v]; }
int bs(int* ptr, int len, int seek) {
  int cur = 0, step = len >> 1;
  if (len == 0) return 0;
  while (step) {
    if (ptr[cur + step] < seek)
      cur += step;
    else
      step >>= 1;
    while (cur + step >= len) step >>= 1;
  }
  if (ptr[cur] < seek) return cur + 1;
  return cur;
}
struct SH {
  int *ptr, len, dl;
  SH(int a = 10) {
    len = a;
    dl = 0;
    ptr = new int[len + 1];
  }
  ~SH() { delete[] ptr; }
  void push(int x) {
    int flg = bs(ptr, dl, x);
    if (flg != dl && ptr[flg] == x) return;
    for (int i = dl; i > flg; --i) ptr[i] = ptr[i - 1];
    ptr[flg] = x;
    dl = min(dl + 1, len);
  }
  void push(SH& sh) {
    size_t nt = min(sh.dl, len);
    for (size_t i = 0; i < nt; ++i) push(sh.ptr[i]);
  }
  void operator=(SH& sh) {
    dl = sh.dl;
    for (int i = 0; i < dl; ++i) ptr[i] = sh.ptr[i];
  }
};
SH dp[100000 + 5][19];
int p[100000 + 5][19];
void LCA(int u, int v, SH& ans) {
  int cur = 0, step = 19 >> 1, np;
  bool find;
  if (u == v) {
    ans.push(dp[u][0]);
    return;
  }
  if (uisa(v, u)) swap(u, v);
  if (uisa(u, v)) {
    find = false;
    while (step) {
      np = p[v][cur + step];
      if (uisa(u, np)) {
        cur += step;
        if (u == np) {
          find = true;
          break;
        }
      } else
        step >>= 1;
      while (cur + step >= 19) step >>= 1;
    }
    ans.push(dp[v][cur]);
    if (!find) LCA(u, p[v][cur], ans);
    return;
  }
  if (uisa(p[u][1], v)) {
    ans.push(dp[u][1]);
    LCA(p[u][1], v, ans);
    return;
  }
  while (step) {
    np = p[u][cur + step];
    if (uisa(np, v))
      step >>= 1;
    else
      cur += step;
    while (cur + step >= 19) step >>= 1;
  }
  ans.push(dp[u][cur]);
  LCA(p[u][cur], v, ans);
}
void Build() {
  vector<int> st;
  for (int i = 1; i <= n; ++i) {
    vis[i] = false;
    p[i][0] = i;
  }
  p[1][1] = 1;
  st.push_back(1);
  t = 0;
  while (!st.empty()) {
    int cur = st.back();
    if (vis[cur]) {
      st.pop_back();
      out[cur] = ++t;
      continue;
    }
    in[cur] = ++t;
    vis[cur] = true;
    for (int i = 2; i < 19; ++i) {
      p[cur][i] = p[p[cur][i - 1]][i - 1];
      dp[cur][i] = dp[cur][i - 1];
      dp[cur][i].push(dp[p[cur][i - 1]][i - 1]);
    }
    for (size_t i = 0; i < eg[cur].size(); ++i) {
      int np = eg[cur][i];
      if (p[cur][1] == np) continue;
      p[np][1] = cur;
      dp[np][1] = dp[np][0];
      dp[np][1].push(dp[cur][0]);
      st.push_back(np);
    }
  }
}
int main() {
  int u, v, a;
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &u, &v);
    eg[u].push_back(v);
    eg[v].push_back(u);
  }
  for (int i = 0; i < m; ++i) {
    scanf("%d", &u);
    dp[u][0].push(i + 1);
  }
  Build();
  while (q--) {
    scanf("%d%d%d", &u, &v, &a);
    SH sh(a);
    LCA(u, v, sh);
    printf("%d", sh.dl);
    for (int i = 0; i < sh.dl; ++i) printf(" %d", sh.ptr[i]);
    printf("\n");
  }
  return 0;
}